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Ice is used as a coolant due to the high specific heat capacities of ice and water and the high specific latent heat of fusion of ice - Leaving Cert Physics - Question 9 - 2021

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Ice is used as a coolant due to the high specific heat capacities of ice and water and the high specific latent heat of fusion of ice. It is the principal coolant us... show full transcript

Worked Solution & Example Answer:Ice is used as a coolant due to the high specific heat capacities of ice and water and the high specific latent heat of fusion of ice - Leaving Cert Physics - Question 9 - 2021

Step 1

What is meant by specific heat capacity?

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Answer

Specific heat capacity is defined as the amount of thermal energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin. It can be expressed mathematically as:

c=QmΔTc = \frac{Q}{m \Delta T}

where:

  • QQ is the energy supplied,
  • mm is the mass of the substance,
  • ΔT\Delta T is the change in temperature.

Step 2

Why does the high specific latent heat of fusion of ice make it a good coolant?

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Answer

The high specific latent heat of fusion of ice makes it a good coolant because it takes a large amount of energy to change ice into water without changing temperature. This property allows ice to absorb a significant amount of heat from the environment, maintaining a low temperature and effectively cooling the surrounding area.

Step 3

Suggest two reasons why the walls of a picnic box are made from hollow plastic rather than solid plastic.

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Answer

  1. Better insulator: Hollow plastic traps air, which is an excellent insulator, reducing heat transfer and keeping the contents cooler for a longer time.
  2. Lighter: Hollow plastic is lighter than solid plastic, making the picnic box easier to carry while still maintaining durability.

Step 4

Calculate the final temperature inside the picnic box when its contents have reached thermal equilibrium.

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Answer

To find the final temperature, we can use the equation:

mcΔT=miLfm_{c} \Delta T = -m_{i} L_{f}

Where:

  • mc=0.25kgm_{c} = 0.25 \, \text{kg} (mass of food)
  • c=17800J kg1K1c = 17800 \, \text{J kg}^{-1} \text{K}^{-1} (heat capacity of food)
  • mi=0.25kgm_{i} = 0.25 \, \text{kg} (mass of ice)
  • Lf=3.3×105J kg1L_{f} = 3.3 \times 10^5 \, \text{J kg}^{-1} (latent heat of fusion of ice)
  • Initial temperature of ice = -18 °C & Initial temperature of food = 10.5 °C.

Now calculating:

  1. For the food:

    ΔT=Tfinal10.5\Delta T = T_{final} - 10.5

  2. For the ice:

    ΔT=0(18)=18\Delta T = 0 - (-18) = 18

Combining both:

0.25imes17800imes(Tfinal10.5)=0.25imes3.3×1050.25 imes 17800 imes (T_{final} - 10.5) = -0.25 imes 3.3 \times 10^5

4450(Tfinal10.5)=825004450(T_{final} - 10.5) = -82500

Tfinal10.5=18.54T_{final} - 10.5 = -18.54

Tfinal=10.518.54T_{final} = 10.5 - 18.54

Calculating gives us:

Tfinal=5.04°CT_{final} = 5.04 \, °C

Step 5

Draw a labelled diagram of a heat pump.

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Answer

The heat pump diagram should include the following components labeled:

  • Compressor
  • Expansion valve
  • Condenser
  • Evaporator

Arrows should indicate the flow of refrigerant through each stage.

Step 6

Explain how a heat pump works.

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Answer

A heat pump operates by transferring heat from one location to another. It consists of:

  1. Evaporator: The refrigerant absorbs heat and evaporates.
  2. Compressor: The gaseous refrigerant is compressed, increasing its temperature.
  3. Condenser: The refrigerant releases heat as it condenses back to liquid.
  4. Expansion valve: The refrigerant expands, dropping in pressure and temperature, returning to the evaporator to repeat the cycle.

Step 7

What observations did the student make?

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Answer

The student might have observed that:

  • The ice melted during the experiment.
  • Heat was transferred from the water to the ice, causing the water temperature to decrease.
  • The rate of melting would vary based on the temperature difference between the ice and the surrounding water.

Step 8

What conclusion could the student have made?

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Answer

The student could conclude that water is a good conductor of heat, as it effectively transfers heat to the ice, facilitating its melting and demonstrating the principles of thermal conductivity.

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